Question

I am using the following PLone + urllib code to proxy responses from another server through a BrowserView

req = urllib2.Request(full_url)

    try:

        # Important or if the remote server is slow
        # all our web server threads get stuck here
        # But this is UGLY as Python does not provide per-thread
        # or per-socket timeouts thru urllib
        orignal_timeout = socket.getdefaulttimeout()
        try:
            socket.setdefaulttimeout(10)

            response = urllib2.urlopen(req)
        finally:
            # restore orignal timeoout
            socket.setdefaulttimeout(orignal_timeout)

        # XXX: How to stream respone through Zope
        # AFAIK - we cannot do it currently

        return response.read()

My question is how could I make this function not to block and start streaming the proxied response through Zope instantly when the first bytes arrive? When interfaces, objects or patterns are used in making streamable Zope responses?

Answer 1

I think there are two ways you can do this. Firstly, the Zope response itself is file-like so you can use the response’s write() method to write successive chunks of data to the response as they come in. Here’s an example where I use a Zope response as a file-like object for a csv.writer.

Or you can use ZPublisher’s IStreamIterators and wrap the response in a ZPublisher.Iterators.filestream_iterator wrapper and return the wrapper.